Figure shows a square loop ABCD with edge length a. The resistance of the wire ABC is r and that of ADC is 2r. The

Q: Figure shows a square loop $ABCD$ with edge length $a$. The resistance of the wire $ABC$ is $r$ and that of $ADC$ is $2r$. The value of magnetic field at the centre of the loop assuming uniform wire is

b (b) According to question resistance of wire $ADC$ is twice that of wire $ABC$. Hence current flows through $ADC$ is half that of $ABC$ i.e. $\frac{{{i_2}}}{{{i_1}}} = \frac{1}{2}$. Also ${i_1} + {i_2} = i$ $==>$ ${i_1} = \frac{{2i}}{3}$ and ${i_2} = \frac{i}{3}$ Magnetic field at centre $O$ due to wire $AB$ and $BC$ (part $1$ and $2$) ${B_1} = {B_2} = \frac{{{\mu _0}}}{{4\pi }}.\frac{{2{i_1}\sin {{45}^o}}}{{a/2}} \otimes $$ = \frac{{{\mu _0}}}{{4\pi }}.\frac{{2\sqrt 2 \,{i_1}}}{a} \otimes $ and magnetic field at centre $O$ due to wires $AD$ and $DC$ (i.e. part $3$ and $4$) ${B_3} = {B_4} = \frac{{{\mu _0}}}{{4\pi }}\frac{{2\sqrt 2 \,{i_2}}}{a}\odot$ Also $i_1 = 2i_2. \,So \,(B_1 = B_2) > (B_3 = B_4)$ Hence net magnetic field at centre $O$ ${B_{net}} = ({B_1} + {B_2}) - ({B_3} + {B_4})$ $ = 2 \times \frac{{{\mu _0}}}{{4\pi }}.\frac{{2\sqrt 2 \, \times \left( {\frac{2}{3}i} \right)}}{a} - \frac{{{\mu _0}}}{{4\pi }}.\frac{{2\sqrt 2 \,\left( {\frac{i}{3}} \right) \times 2}}{a}$ $ = \frac{{{\mu _0}}}{{4\pi }}.\frac{{4\sqrt 2 \,i}}{{3a}}(2 - 1)\, \otimes = \frac{{\sqrt 2 \,{\mu _0}i}}{{3\pi \,a}} \otimes $

SECTION - A [PHYSICS - MCQ]

GSEB - English Medium

Figure shows a square loop $ABCD$ with edge length $a$. The resistance of the wire $ABC$ is $r$ and that of $ADC$ is $2r$. The value of magnetic field at the centre of the loop assuming uniform wire is

A$\frac{{\sqrt 2 \,{\mu _0}i}}{{3\pi \,a}}\odot$
B$\frac{{\sqrt 2 \,{\mu _0}i}}{{3\pi \,a}} \otimes $
C$\frac{{\sqrt 2 \,{\mu _0}i}}{{\pi \,a}}\odot$
D$\frac{{\sqrt 2 \,{\mu _0}i}}{{\pi \,a}} \otimes $

Diffcult

STD 11 Science
NEET
STD 12 - 4. Moving charges and magnetism
Physics

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